Problem: Simplify the following expression and state the condition under which the simplification is valid. $n = \dfrac{8y^3 - 16y^2 + 8y}{-y^2 + 4y - 3}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ n = \dfrac {8y(y^2 - 2y + 1)} {-1(y^2 - 4y + 3)} $ $ n = -\dfrac{8y}{1} \cdot \dfrac{y^2 - 2y + 1}{y^2 - 4y + 3} $ Next factor the numerator and denominator. $ n = - 8y \cdot \dfrac{(y - 1)(y - 1)}{(y - 1)(y - 3)}$ Assuming $y \neq 1$ , we can cancel the $y - 1$ $ n = - 8y \cdot \dfrac{y - 1}{y - 3}$ Therefore: $ n = \dfrac{ -8y(y - 1)}{ y - 3 }$, $y \neq 1$